8th August

Question 5iv

Find the cartesian equation of the path of the mid-point of (2+2cos⁡t,tan⁡t){(2+2\cos t, \tan t)} and (−2,0){(-2,0)}.

Naturally, we find the coordinates of the mid-point: (cos⁡t,tan⁡t2).{(\cos t, \frac{\tan t}{2}).} To convert to the cartesian equation, we need to introduce xx and yy. This means that

x=cos⁡t,y=tan⁡t2 x = \cos t, \quad y = \frac{\tan t}{2}

We now need to combine the two into one equation. For trigonometric expressions where it is not ideal to make tt the subject, we can use our Pythagorean trigonometric identities: the tan⁡2A+1=sec⁡2A\tan^2 A + 1 = \sec^2 A formula is most applicable here.

x=cos⁡t{x = \cos t} means that sec⁡t=1x{\sec t = \frac{1}{x}}. Substituting this and tan⁡t=2y{\tan t = 2y} into the trigonometric identity,

4y2+1=1x2 4y^2 + 1 = \frac{1}{x^2}

which is our cartesian equation (the answer your school provided made yy the subject from here).

Question 4iv

Find the area of the region bounded by the curve, the xx-axis and the normal in (iii)

Curve: x=3t−t3,y=t2−1t,t≥1{x = 3t-t^3, \quad y = t^2 - \frac{1}{t}, \quad t \geq 1}.

Line (normal at t=2{t=2}): 34y=72x+263{34y = 72x+263}

Question 5iv

Find, in radians, the acute angle between the tangent and the normal

Recall our discussion recently about the relationship between the angle and the gradient: tan⁡θ=gradient{\tan \theta = \textrm{gradient}}, where θ\theta is the angle the line makes with the xx-axis.

Turns out the normal has equation y=ln⁡2{y=\ln 2} so it is horizontal and behaves like the xx-axis. tan⁡θ=−116{\tan \theta = -\frac{1}{16}}. We want just an acute angle so our lines are infinite so θ=tan⁡−1116{\theta = \tan^{-1} \frac{1}{16}} (taking just the positive value) will give us the answer.